--- title: "LT loss and voltage drop, 60 kW four feeders" subject: "Transmission and Distribution System Design" chapter: 7 key: "semester_8/transmission_and_distribution_system_design:lt-feeders-60kw-1-5km-380v" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/lt-feeders-60kw-1-5km-380v site: BCT Engineering --- # LT loss and voltage drop, 60 kW four feeders Transmission and Distribution System Design · Chapter 7: Distribution System Design - Asked in: 2067 Chaitra (old course), 12 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/lt-feeders-60kw-1-5km-380v - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7 ## Question A particular load center has a peak demand of 60 kW at a power factor of 0.9. The distribution transformer has four 3-phase LT feeders at 380 V of equal length of 1.5 km each. If the per phase resistance and reactance of LT conductors are 0.9 Ω/km and 0.25 Ω/km, determine the following: i) LT percentage peak power loss ii) Maximum percentage voltage drop in LT. Assume uniformly distributed load along the feeders. ## Short answer Each feeder carries 25.32 A. Total three-phase loss = 4 × (25.32² × 1.35) = 3462.6 W, 5.77% of 60 kW. Voltage drop per phase = ½ × 25.32 × 0.919 × 1.5 = 17.45 V, 7.96%. ## Answer #### Formulas (uniformly distributed load, 3-phase) For each feeder of length $l$ carrying sending-end current $I$: $$ \begin{aligned} P_L &= 3\cdot\frac13 I^2 r l = I^2 r l\\ V_{D,ph} &= \frac12 I(r\cos\phi + x\sin\phi)\,l\\ \%V_D &= \frac{V_{D,ph}}{V_L/\sqrt3}\times100 \end{aligned} $$ The peak load is shared equally between the feeders. #### Data | Quantity | Value | |---|---| | Peak demand | 60 kW | | Power factor | 0.9 ($\sin\phi = 0.4359$) | | Feeders | 4, each 1.5 km | | LT voltage | 380 V | | r, x | 0.9, 0.25 Ω/km | #### Current per feeder $$ P_{feeder} = \frac{60}{4} = 15\ \text{kW},\quad I = \frac{15\,000}{\sqrt3\times380\times0.9} = 25.32\ \text{A} $$ #### i) LT percentage peak power loss Resistance per phase per feeder $R = 0.9\times1.5 = 1.35\ \Omega$. $$ \begin{aligned} P_{L,feeder} &= I^2R = 25.32^2\times1.35 = 865.65\ \text{W}\\ P_{L,total} &= 4\times865.65 = 3462.6\ \text{W}\\ \%P_L &= \frac{3462.6}{60\,000}\times100 = 5.77\ \% \end{aligned} $$ #### ii) Maximum percentage voltage drop $$ \begin{aligned} r\cos\phi + x\sin\phi &= 0.9(0.9) + 0.25(0.4359) = 0.919\ \Omega/\text{km}\\ V_{D,ph} &= \frac12\times25.32\times0.919\times1.5 = 17.45\ \text{V}\\ V_{ph} &= 380/\sqrt3 = 219.39\ \text{V}\\ \%V_D &= \frac{17.45}{219.39}\times100 = 7.96\ \% \end{aligned} $$ Line-to-line drop $= \sqrt3\times17.45 = 30.23$ V. #### Why the UDL factors - The current falls linearly from $I$ at the transformer to zero at the end, so the drop equals that of the full current flowing over half the length ($\frac12 Il$). - Loss integrates $I^2$, giving one-third of the lumped value; for three phases, $3\times\frac13 I^2 r l = I^2 r l$. The drop (≈ 8 %) is at the upper end of the usual LT limit, and the loss (≈ 5.8 %) is acceptable but could be reduced with a larger conductor. **Answer: i) LT peak power loss = 5.77 % ii) Maximum voltage drop = 7.96 %.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For 60 kW over 4 feeders at 380 V, p.f. 0.9, the current per feeder is about: A. 25.32 A B. 101 A C. 14.6 A D. 50.6 A Answer: A. I = 15000/(sqrt3 x 380 x 0.9) = 25.32 A. (Key point: Feeder current) 2. The total % peak loss comes from: A. I^2 R for one feeder over 60 kW B. 4 x I^2 R (R = r l = 1.35 Ω) over 60 kW C. 3 x I^2 R over 15 kW D. 4 x I^2 x l over 60 kW Answer: B. UDL loss per feeder is I^2 R; four feeders give 3462.6 W = 5.77 %. (Key point: Loss calculation) 3. Which factor multiplies I and l in the per-phase drop for this feeder? A. r + x B. r cosφ + x sinφ = 0.919 Ω/km, with 1/2 for UDL C. r sinφ + x cosφ D. sqrt(r^2 + x^2) without 1/2 Answer: B. V_D = 1/2 x I x (r cosφ + x sinφ) x l = 17.45 V. (Key point: Drop calculation) 4. Why is the drop 1/2 I l while loss uses 1/3? A. Drop integrates I (linear) giving 1/2; loss integrates I^2 giving 1/3 B. Drop has two phases C. Loss is for three phases only D. They are arbitrary conventions Answer: A. Current falls linearly: integral of (1-x/l) is l/2 and of its square is l/3. (Key point: UDL factors reason) ## Close variants set in other exams - [LT loss and voltage drop, four 2 km feeders](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Alt-feeders-40kw-2km-380v) — Transmission and Distribution System Design, chapter 7, asked 2× - [LT feeder options: loss and voltage drop, 50 kW](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Alt-feeder-two-options-50kw) — Transmission and Distribution System Design, chapter 7, asked 3× ## Related in Transmission and Distribution System Design - [Distribution loss and voltage drop, four LT feeders](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aload-center-40kw-four-lt-feeders) — Transmission and Distribution System Design, chapter 7 - [Peak loss and voltage drop for LT feeder options](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Alt-feeder-configuration-options) — Transmission and Distribution System Design, chapter 7, asked 2× - [Peak demand of load center from load pattern](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Apeak-demand-load-center-load-pattern) — Transmission and Distribution System Design, chapter 6, asked 2× - [Loss and voltage drop, 16 km uniformly varying load](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Auniformly-varying-load-16km-feeder) — Transmission and Distribution System Design, chapter 7